Question

Arătați că nr
p=(√5+√2)2-√2(√5+√2)-√5(√2-2√5) ∈N

Answer 1

p=(√5+√2)²-√2(√5+√2)-√5(√2-2√5)=

=5+2√10+2-√10-2-√10+20=

=5+2-2+20=25

Answer 2

$p=( \sqrt{5} + \sqrt{2} )^2- \sqrt{2} ( \sqrt{5} + \sqrt{2} )- \sqrt{5} ( \sqrt{2} -2 \sqrt{5} ) \\ p= \sqrt{5}^2 + 2 \cdot \sqrt{5} \cdot \sqrt{2} + \sqrt{2} ^2- \sqrt{10} -2- \sqrt{10} +10 \\ p=5+2 \sqrt{10} +2- \sqrt{10} -2- \sqrt{10} +10 \\ p=15 \in N$